Dynamic Renormalization group in Momentum space

Question

I am trying to reproduce the computations of Appendix B of Fractal Concepts on Surface Growth (Barabási & Stanley), about the computation of the critical exponents of KPZ equation using Dynamic Renormalization Group techniques. I have the KPZ equation,

$$\frac{\partial h\left(\vec{x},t\right)}{\partial t}=v\nabla^{2}h+\frac{\lambda}{2}\left(\nabla h\right)^{2}+\eta\left(\vec{x},t\right)$$

where $\eta$ is a Gaussian white noise with correlations $\left\langle \eta\left(\vec{x},t\right)\eta\left(\vec{x}',t'\right)\right\rangle =2D\delta\left(\vec{x}-\vec{x}'\right)\delta\left(t-t'\right)$.

Then, I do a perturbative expansion and solve the integrals to renormalize $\nu$. I have no problem with that. I get the result indicated on the book,

$$\tilde{\nu}=\nu\left[1-\frac{\lambda^{2}DK_{d}\left(d-2\right)}{4d\nu^{3}}\int_{0}^{\Lambda}q^{d-3}dq\right]$$

where the integral is divergent for $d\leq d_c = 2$ as $q \rightarrow 0$. In section B.3, the book tells how to renormalize the integral in $dq$, and it gives two steps:

1) The equivalent of coarse graining in real space, which to integrate in a momentum shell $\Lambda /b \leq q \leq \Lambda$, leaving the integral $0 \leq q \leq \Lambda / b$ untouched. We will use also $b=e^\ell\simeq(1+\ell)$.

2) A rescaling, which in space is $x\rightarrow bx$ and in momentum is $q \rightarrow q/b$.

However I don't understand how this is done in the divergent integral. What you can see in the book is that they compute

$$\int_{\Lambda/b} ^{\Lambda}q^{d-3}dq \simeq \int_{\Lambda (1-\ell)} ^{\Lambda}q^{d-3}dq \simeq \ell \Lambda^{d-2}$$

where they let $\Lambda=1$ without loss of generality. Then they say that this result is the one we get when we are with long wavelenghts (slow modes) only,

$$\nu^<=\nu\left[1-\ell\frac{\lambda^{2}DK_{d}\left(d-2\right)}{4d\nu^{3}}\right]$$

And then they apply the rescaling as $\tilde{\nu} =b^{z-2}\nu^< \simeq \nu^< [1+\ell(z-2)]$. Then the book substitutes the expression of $\nu^<$ and operates to order $\mathcal{O}(\ell)$ to find the flow equations for the parameter.

Under my point of view, what it is doing is to integrate the fast modes and put this result inside the expression of $\nu^<$, which are the slow modes, and forgets about the integral between 0 and $\Lambda/b$.

So my question is: I don't understand why we integrate over the fast modes and then substitute directly in the expression of $\nu$. More precisely, I don't know what happened to the integral between 0 and $\Lambda/b$. What am I missing here?.

I tried to see why this is done in this way. I tried to separe the integral in slow and fast modes:

$$\int_{0} ^{\Lambda}q^{d-3}dq= \int_{0} ^{\Lambda/b}q^{d-3}dq + \int_{\Lambda/b} ^{\Lambda}q^{d-3}dq$$

Then the second integral can be done and it is a constant. As I see, when I rescale $q\rightarrow q/b$, the first integral has the limit $\Lambda/b \rightarrow \Lambda /b^2$. In addition to that, the constant will depend also on $\Lambda$, so it will be also rescaled -unless we use the trick to put $\Lambda = 1$ as they do, so I am really not sure on how this works.

So, how this separation into slow and fast modes happen? Any explanation and/or useful sources are welcome.

UPDATE: I added a bounty for getting more attention into this question. In addition to that, I want to point out that I've read that in fact all this come from the beginning, so I have to split the field $h(\vec{k},\omega)=h^<(\vec{k},\omega)+h^>(\vec{k},\omega)$ and then average only over the fast modes $h^>(\vec{k},\omega)$. This will give the expression for $\nu ^<$ I am trying to find, after doing to one-loop integral, that will be only in the momentum shell. However I couldn't get a lot of detail on how to do this process. Thank you!

Answer 1

The OP's problem seems to come from a slight misunderstanding of the RG procedure. Not having the OP's reference available, I will comment on the standard $\phi^4$ theory, which should be sufficient to answer the problem, the action of which is $$S=\int_x\left[ \frac{(\nabla \phi)^2}{2}+\frac{r}{2}\phi^2+\frac{u}{4!}\phi^4\right].$$

The coupling $\nu$ seems to be equivalent to the mass term $r$, so I will only discuss the renormalization of this term.

The idea of Wilson RG is to integrate out high-momenta modes $\phi_>$ while leaving untouched the slow modes $\phi_<$. The partition function is written as $$Z=\int D\phi_< \int D\phi_> e^{-S[\phi_< ,\phi_>]}.$$

Then $S[\phi_< ,\phi_>]$ is split into $S_<[\phi_<]$ which involves only slow modes, and $S_{int}[\phi_< ,\phi_>]$ which involves the interaction between the slow and the fast modes. After integrating out the fast modes, of momentum $\Lambda/b <q<\Lambda$, one obtains a new action for the slow modes, $$Z=\int D\phi_< e^{-S_{<,b}[\phi_<]},$$ with $$S_{<,b}[\phi_<]=S_<[\phi_<]-\log\left[\int D\phi_> e^{-S_{int}[\phi_< ,\phi_>]}\right].$$ Notice that by definition, the correction to $S_<[\phi_<]$ coming from the log can only involve loops with integrals over momenta $q\in[\Lambda/b ,\Lambda]$.

Performing a loop expansion of the log term, we obtain a quadratic term in the slow modes of the form ($\alpha$ is some constant which we could figure out) $$\alpha\int_xu\int_{\Lambda/b <q<\Lambda}\langle\phi_>(q)\phi_>(-q)\rangle_{>,0}\phi_<(x)^2\equiv\int_x\delta r_b\phi_<^2(x),$$ which means that the quadratic term of $S_{<,b}[\phi_<]$ is thus $\int_x r_b\phi_<^2(x),$ with $$r_b=r+\delta r_b =r+\alpha\int_xu\int_{\Lambda/b <q<\Lambda}q^{d-3},$$ assuming $\langle\phi_>(q)\phi_>(-q)\rangle_{>,0}\propto q^{-2}$.

We thus see that the renormalized mass term (before rescaling) involves only integrals over high momenta (which is the point of Wilson procedure).